${\mathrm{KMnO}}_4$ (purple) is reduced to ${\mathrm{K}}_2{\mathrm{MnO}}_4$ (green) by ${\mathrm{SO}}_3^{2-}$ in basic medium. 1 mole of ${\mathrm{KMnO}}_4$ is reduced by

Multiply by change in oxidation number

$$\begin{gathered} 2{\mathrm{MnO}}_4^{-} + {\mathrm{SO}}_3^{2-} \stackrel{ }{\to }{\mathrm{SO}}_4^{2-} + 2{\mathrm{MnO}}_4^{2-} \\ \mathrm{Thus}, 2{\mathrm{MnO}}_4^{2-} \equiv 1 {\mathrm{SO}}_3^{2-} \\ 1 {\mathrm{MnO}}_4^{-} \equiv 0.5 {\mathrm{SO}}_3^{2-} \end{gathered}$$